# Tag Info

2

Since $T - \lambda I$ is also normal, we have $$\| T - \lambda I \| = \text{spr} (T - \lambda I) = 0,$$ showing that $T = \lambda I$. (I recently asked basically the same question (Self-adjoint operator with single point spectrum), but your formulation is more general so I thought it might be worth sharing the answer here.)

1

using spectral theorem since $T$ is normal it exist a spectral measure $E$ such that $$Tx=\int_{\sigma(T)}tdE(x)=\int_{\{\lambda\}}tdE(x)=\lambda E(\{\lambda\})(x)=\lambda E(\sigma(T))(x)=\lambda I (x)=\lambda x$$

-2

completeness in hilbert space means that the metric associated with inner product converges,let draw a 1/x as a function on a graph,you will see its area converges to zero. as x approaches to infinity 1/x goes to zero

3

Yes. For a positive linear operator $A$, $\|A\| = \sup \{(Ax, x): \; \|x\| = 1\}$. If $A \ge B$, $(Ax, x) \ge (Bx, x)$, so we must have $\|A\| \ge \|B\|$.

0

For any subset: $$A\subseteq\mathcal{H}:\quad A\cap A^\perp=\{0\}$$ As well as it is: $$A\subseteq\mathcal{H}:\quad\overline{\langle A\rangle}^\perp=A^\perp=\overline{\langle A^\perp\rangle}$$ Moreover it holds: $$A\subseteq\mathcal{H}:\quad A^{\perp\perp}=\overline{\langle A\rangle}$$

-1

1) is just Pythagoras theorem applied to $f(x)$ 2) is just that you need completness because somewhere you use a cauchy sequence you need to converge 3)What's your definition of $Tr(f)$ ?

2

1) You know that $$f(x)=\sum_{i=1}^\infty \lambda_ie_i$$ where $\lambda_i\in\mathbb R$. Now, since $\{e_i\}$ is an orthonormal basis, you'll get, $$\left< f(x),e_i\right>=\sum_{i=1}^\infty \left<\lambda_j e_j,e_i\right>=\sum_{j=1}^\infty \lambda_j\underbrace{\left<e_j,e_i\right>}_{=\delta_{ij}}=\lambda_i.$$ Notice that $\left<f(x),e_i\... 0 I'm not following you here. So let$H$be a Hilbert space and$V$a (not necessarily closed) subspace, then$V^{\perp}=\left\{w\in H\mid \left\langle v,w\right\rangle=0 \mbox{ for all }v\in V\right\}$is the orthogonal complement. Using the continuity of the inner-product you can show that$V^{\perp}$is a closed subspace of$H$. It follows that$V^{\perp\...

2

Start with $$\overline{\mathcal{R}(T)}=\mathcal{N}(T)^\perp.$$ As you noted, $\mathcal{N}(T^{1/2})\subseteq\mathcal{N}(T)$. Because $T^{1/2}$ is selfadjoint, $$\|T^{1/2}x\|^2=(T^{1/2}x,T^{1/2}x)=(Tx,x).$$ Therefore $\mathcal{N}(T)\subseteq\mathcal{N}(T^{1/2})$. So, $$\overline{\mathcal{R}(T)}=\mathcal{N}(T)^{\perp}=\... 2 Of course as Hilbert spaces L^2(S^1) and L^2\bigl([0,1]\bigr) are isomorphic, and you could also say that L^2\bigl([0,1]\bigr) is the prime example of a Hilbert space arising from Lebesgue theory. But note that L^2(S^1) is one of the most important Hilbert spaces in the world, and there definitively is an essential difference between L^2\bigl([0,1]\... 3 L^2 spaces should not be sensitive to the topology or shape of whatever underlying space you're working with. Indeed, given a "manifold" (a generalization of circles and surfaces), one way of defining an L^2 space on it is to pick a chart D^n \to M, where D is the unit disc, that is injective except at a set of measure zero, and then pull back ... 1 The "basis" you imply in 4. and 5. is the set of all shifted delta distributions \{\delta_x (f) = f(x)\,\forall\,f\in \mathscr{S}(\mathbb{R}^N)\} where \mathscr{S}(\mathbb{R}^N)\subset L^2(\mathbb{R}^N) is the Schwartz Space of bounded smooth functions f:\mathbb{R}^N\to\mathbb{R}^N such that any derivative D^\alpha (p\,f) of any multiple f p of f ... 3 The `basis' you described in 3 is given by the Dirac measures, and it is not in L^2, meaning that \langle x|x\rangle is not defined. In other words, states concentrated at a single point is not allowed in the theory from a strict mathematical point of view. So 4 is not true. If you want to include states like |x\rangle there is a formalism called ... 0 There are two mathematically rigorous (and quite general) Hilbert Space formalisms you might be looking for. Both can be seen as attempts to salvage the engine of Dirac's original bra-ket algorithm, while avoiding its mathematical embarrassments. The first - created by von Neumann - replaces Dirac's kets by vectors in an abstract Hilbert space \mathscr{H},... 1 Without orthogonality this is false: an example is given by Robert Israel. Orthogonality implies that \|u+v\|^2 = \|u\|^2+\|v\|^2 for u\in M, v\in N. Thus, if a sequence (u_n+v_n) converges, the inequalities such as$$\|u_n-u_m\|\le \|(u_n+v_n)-(u_m+v_m)\|$$imply convergence of both u_n and v_n. So, u_n\to u\in M and v_n\to v\in N, which ... 0 Recall Gelfand's formula for the spectral radius of a bounded operator T:$$r(T) = \lim_{n\to\infty} \|T^n\|^{\frac1n}. $$If T is a self-adjoint operator and \|f\|=1 then$$\|Tf\|^2 = \langle Tf, Tf\rangle = \langle T^2f, f\rangle\leqslant\|T^2f\|\|f\|=\|T^2f\| $$which implies \|T^2\|=\|T\|^2. By induction it follows that \|T^{2^n}\|=\|T\|^{2^n} ... 0 Please see the following rather self-contained proof: Link: http://people.math.gatech.edu/~heil/6338/summer08/section5a_adjoint.pdf 4 Here is a hint: show that K_u(R_0) is both closed and convex. Then apply a result your professor proved - or is in your book - about closed and convex sets to conclude that K_u(R_0) contains a unique point that is closest to the origin in H. 1 (1). For a Banach space B, with scalars S=\mathbb R or S=\mathbb C, the weak^* topology on B^* is defined to be the weakest topology on the set B^* such that for every x\in B the function$$\bar x:B^*\to S$$is continuous, where$$\bar x (f)=f(x).$$So it is necessary and sufficient that \bar x^{-1}U=\{f\in B^*:f(x)\in U\} is a weak^*-open ... 2 I suggest the following link http://mathforum.org/library/drmath/view/74532.html. One can construct the hyperplane that bisects the two vectors, then the matrix U is the reflection with respect to this hyperplane. For the given problem, x-y is perpendicular to this hyperplane. One can show this by (x-y)\cdot(x+y)=0. Then for v=(x-y)/||x-y||, the ... 2 If you are just looking for existence, consider the span S = \operatorname{span} \{x,y\}, take two orthogonal vectors u_1,u_2\in S, and using Gram-Schmidt extend this set to an orthogonal basis u_1,u_2,\ldots,u_n of \mathbb R^n. Now Consider the matrix$$U=\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\\&&1\\&&&...

1

$\newcommand\ip[2]{\langle#1,#2\rangle}$ It's false. For example define $T:\Bbb C^2\to\Bbb C^2$ by $$T(x_1,x_2)=(x_2,0);$$then $||T||=1$ while the AM-GM inequality followed by Holder's inequality shows that $$|\ip {Tx}x|\le\frac{||x||^2}{2}.$$ It should probably be noted that it's less trivially false in the complex case than in the real case. As has been ...

0

Let $x,y \in X$, then we have \begin{align*} \sum_{k=0}^3 i^k B(x+i^k y, x+i^ky) &= B(x+y, x+y) - B(x-y,x-y) \\ & \quad + i \left[B(x+iy,x+iy) - B(x-iy, x-iy) \right] \\ &= 2 B(y,x)+2B(x,y) + 2B(x,y)-2B(x,y) + 2 B(y,x) \\ &= 4B(y,x). \end{align*} Since $B(x+i^k y, x+i^ky) \in \mathbb{R}$, we have \begin{align*} 4\overline{B(y,x)} &=\...

0

Let $\lambda\notin \sigma(A)$ then $A-\lambda 1$ is invertible, denote it's inverse by $C\in B(H_2)$. We have $C(A-\lambda1)=(A-\lambda1)C=1$. Therefore, $D=U^{-1}CU\in B(H_1)$ is the inverse of $(B-\lambda1)$ (why? write $B=U^{-1}AU$). Thus, $\lambda \notin \sigma(B)$. Now, if $\lambda\in \sigma_p(A)$ that means $\exists v\ne 0$ such that $Av=\lambda v$. ...

1

You are almost done. Since $f_n\to f$ pointwise, and $f_n'\to h$ in $L^2$, it follows that for every $a,b\in\mathbb{R}$, $$f(b)-f(a)= \lim_{n\to\infty}(f_n(b)-f_n(a)) = \lim_{n\to\infty}\int_a^b f_n'(t)\,dt = \int_a^b h(t)\,dt$$ The last step here is based on the Cauchy-Schwarz inequality, $$\int_a^b |f_n'(t)-h(t)|\,dt \le (b-a)^{1/2}\left(\int_a^b |f_n'(t)-... 2 For every x we have x=Px+(x-Px) where Px\in \operatorname{im}P and (x-Px)\in \ker P. So, \ker P is a complement of \operatorname{im} P. Suppose it is not an orthogonal complement; then there exist u\in \operatorname{im}P and v\in \ker P such that \operatorname{Re}\langle u, v \rangle < 0. For sufficiently small t>0 we have \|u+tv\... 1 The GNS construction doesn't have a canonical way of choosing "the right amount" of states. For an arbitrary algebra one uses all states (see for instance Corollary I.9.11 in Davidson's C^*-Algebras by Example). But this is often way too much. Let's think of the easiest concrete example: let A=M_2(\mathbb C). If we follow the abstract recipe, we need ... 1 Without loss of generality, assume that \lambda_k \ne 0 for all k. Otherwise, discard v_k from your orthonormal set, and discard the term \lambda_k (x,v_k)v_k from the defining sum for F. Nothing is changed by making such changes. And assume that \{ v_k \} is an orthonormal set by discarding 0 vectors and renormalizing. The largest domain \... 2 Assuming \{ e_n \} and \{ e_n'\} are orthonormal, define$$ Lx = \sum_{n=1}^{\infty}(x,e_n)e_n' $$The operator L is linear and satisfies$$ \|Lx\|^2=\sum_{n=1}^{\infty}|(x,e_n)|^2 \le \|x\|^2. $$Now assume that \{e_n\} is a complete orthonormal basis, and assume that d=\left(\sum_{n=1}^{\infty}\|e_n-e_n'\|^2\right)^{1/2} ... 1 Pedestrian way.. Eigenvector:$${\sum}_k(\lambda-\lambda_k)\alpha_kv_k=\lambda({\sum}_k\alpha_kv_k)-{\sum}_k\alpha_k(\lambda_kv_k)=\\=F({\sum}_k\alpha_kv_k)-{\sum}_k\alpha_k(Fv_k) =F({\sum}_k\alpha_kv_k-{\sum}_k\alpha_kv_k)=F(0)=0$$Nullvector:$${\sum}_k|\lambda-\lambda_k|^2|\alpha_k|^2=\|{\sum}_k(\lambda-\lambda_k)\alpha_kv_k\|^2=\|0\|=0The rest ... 1 This can be done by a straightforward calculation: For all x \in X and y \in Y we have \begin{align*} \langle x, S(y) \rangle &= \langle x, J_X^{-1}(T'(J_Y(y))) \rangle = \langle x, J_X^{-1}(T'( \langle y, - \rangle)) \rangle = \langle x, J_X^{-1}(\langle y, T(-) \rangle) \rangle \\ &= \overline{\langle J_X^{-1}(\langle y, T(-) \... 1 This operator is the orthogonal projection onto U, it holds C^2=C: First we find that \langle Cv,e_n\rangle = \langle v,e_n\rangle, $$since (e_n) is orthonormal. This implies$$ C^2v = \sum_n \langle Cv,e_n\rangle Ce_n = \sum_n \langle v,e_n\rangle e_n =Cv. $$Hence, C^2 = C, which implies C^{1/2}=C. Edit: I assumed that the scalar product ... 0 You should know that for each f \in L^{p}_{loc}(\Omega) we can define a Distribution by: T_{f}: \mathcal{D}(\Omega) \rightarrow\mathbb{R}, where T_{f}(\phi) = \displaystyle\int_{\Omega}f\phi. In the case, T_{f} is called regular distribution. The Du Bouis Reymond Lemma states that the map f\mapsto T_{f} is one-to-one. However, there are ... 1 The fact that \iota is HS means that if \{e_n\} is an orthonormal basis of U, then$$\tag{1} \sum_n\langle e_n,e_n\rangle_V=\sum_n\|\iota(e_n)\|_V^2<\infty. $$After identifying U and V with their respective duals, we have C:V\to U given by$$\tag{2} \langle Cv,u\rangle_U=\langle v,\iota (u)\rangle_V=\langle v,u\rangle_V. In particular, ... 2 You can write unravel this definition by writing \begin{align} Af(x) & = \int_{0}^{x}K(x,y)f(y)dy+\int_{x}^{1}K(x,y)f(y)dy \\ & =\sinh(1-x)\int_{0}^{x}\sinh(y)f(y)dy+\sinh(x)\int_{x}^{1}\sinh(1-y)f(y)dy. \end{align} This is a typical kind of Green function expression. First note that (Af)(0) = 0,\;\;\; (Af)(1)=0. $$Then, \begin{... 2 Weak convergence means: for every f the sequence \{f(x_n)\} tends to 0. When you are proving a statement that begins with "for every f", you don't get to choose f. So, there is no reason for \|f\| to be small. Your argument doesn't work. Instead, use Bessel's inequality \sum |\langle y_f, x_n\rangle|^2 \le \|y_f\|^2, and the fact that the ... 1 Starting with the half of the question that no longer exists: M invertible does not imply T invertible. Say \phi=e_{-2}. Then M is invertible. But if f=e_1 then Tf=P(e_{-1})=0, so T is not invertible. On the other hand T invertible does imply M invertible. Suppose M is not invertible. Let \epsilon>0, and set$$E=\{x:|\phi(x)|<\...

2

$W_C = \{u + iv \in X_C \mid u, v \in W\}$ is invariant under $T_C$, so we may as well forget about the real Banach spaces and consider an operator $T$ on a complex Banach space $X$ with a finite-codimensional invariant subspace $W$, such that $\sigma(T) \subseteq \mathbb R$. The claim is that $\sigma(T|_W) \subseteq \mathbb R$. If $\lambda \notin \sigma(T)... 0 No, that's not a good definition of an orthogonal basis. It's not enough to say every vector can be written as a linear combination, because you may also need infinite sums, which are no longer linear combinations. The sentence "other vectors not in the Hilbert space may also be written as a linear combination of these vectors." makes no sense. What are ... 2 Looking at the definition of a norm on wikipedia (which matches the definitions I encountered in textbooks) (link here) if$V$is a vector space, then a norm is a function$\rho: V\to \mathbb R$that satisfies certain properties so it is quite explicitly stated that the norm of every vector is a real number, thus finite. 1 This is exactly the uniqueness in the polar decomposition. You have, since$v $is a partial isometry, $$\tag {2}{\text {ran}\,v^*v}= {\text {ran}\,v^*}=(\ker v)^\perp=(\ker y)^\perp=\overline {\text {ran}\,y}.$$ Suppose that$w,z $gives another such decomposition of$x $. Let$p=v^*v=w^*w $. Then, since$py=y$, we have $$y^2=y^*y=y^*py=y^*v^*vy=x^*x=|x|... 1 Any bounded sequence in a separable Hilbert space (which is reflexive) has a weakly convergent subsequence. Added on edit: See Theorem 3.18 of the same book. 0 Hint: use the Riesz theorem: Let y\in R, the linear function defined on H by f_y(x)=<T(x),y> is continuous, thus you can find T^*(y)\in H such that f_y(x)=<x,T^*(y)>. -1 This is (tightly related to) formulae (6.11.2) in Maz'ya's book on Sobolev spaces, 2011 edition; or Corollary 1 in §4.11.1 in the 1985 edition. 1 Say x \in V. Since the e_n are a basis for H and x \in H, x = \sum_n c_n e_n for some coefficients c_n \in \mathbb{K}. Since Q and 1-Q are orthogonal projections whose sum is 1, x = Qx + (1-Q)x = (Q \sum_n c_n e_n) + ((1-Q) \sum_n c_n e_n). But the second term is zero since x \in V implies Qx = x. So we have x = Q \sum_n c_n e_n, and ... 1$$T^2x=\sum_n\lambda_n\langle Tx,\phi_n\rangle\phi_n=\sum_n\lambda_n\left\langle \sum_m \lambda_m\langle x,\phi_m\rangle\phi_m,\phi_n\right\rangle \phi_n\\ =\sum_n\lambda_n\sum_m\lambda_m\langle x,\phi_m\rangle\langle\phi_m,\phi_n\rangle \phi_n=\sum_n\lambda_n\sum_m\lambda_m\langle x,\phi_m\rangle||\phi_n||^2\delta_{m,n} \phi_n\\ =\sum_n\lambda_n\lambda_n\... 2 Usually,$(x_k)_{k \in \Bbb N}$is called an orthogonal sequence if$\langle x_j, x_k \rangle = 0$whenever$j \neq k$. 1 Continuous: Note that $$\|Au\|^2 = \sum_k \left(\sum_h a_{k,h} u_h\right)^2 \leq \sum_k \left(\left(\sum_h a_{k,h}^2\right) \|u\|^2\right)$$ Not Compact: Note that$(Ae_n)_k = a_{k,n}$. Thus, we note that for any$n$, we have$(Ae_n)_{n-1} = 1$, so that$\|Ae_n\| \geq 1$. Conclude that$A e_n \not \to 0$. 0$(3)$must not hold for$u=y-v$(only if$y\in U$). That was my mistake. Let$U$and$H$be Hilbert spaces and$\iota$be an embedding of$U$into$H$. Then, $$\pi x:=u\;\;\;\text{for }x\in H\text{ with }x=\iota u+y\text{ for some }u\in U\text{ and }y\in\left(\iota U\right)^\perp$$ is a well-defined mapping$H\to U$. If$\iota$is an isometry, then$\$\...

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